Algebra DPP 1.3 Quadratics

Q1. For what values of b do the equations:

1988x² + bx + 8891 = 0 and 8891x² + bx + 1988 = 0 have a common root?

Solution:

We are given two quadratic equations:

$1988x^2 + bx + 8891 = 0$
$8891x^2 + bx + 1988 = 0$

We are to find the values of $b$ for which they have a common root.

Step 1: Let $\alpha$ be the common root.

Then $\alpha$ satisfies both equations. So,

From equation (1):

$1988\alpha^2 + b\alpha + 8891 = 0 \quad \text{(i)}$

From equation (2):

$8891\alpha^2 + b\alpha + 1988 = 0 \quad \text{(ii)}$

Step 2: Subtract equation (i) from equation (ii):

$(8891\alpha^2 - 1988\alpha^2) + (b\alpha - b\alpha) + (1988 - 8891) = 0$ $(8891 - 1988)\alpha^2 - (8891 - 1988) = 0$ $6903\alpha^2 - 6903 = 0$ $\alpha^2 = 1 \Rightarrow \alpha = \pm 1$

Step 3: Plug back $\alpha = 1$ and $\alpha = -1$ into either equation to find corresponding $b$ .

Case 1: $\alpha = 1$
Plug into equation (1):

$1988(1)^2 + b(1) + 8891 = 0 \Rightarrow 1988 + b + 8891 = 0 \Rightarrow b = -10879$

Case 2: $\alpha = -1$

$1988(-1)^2 + b(-1) + 8891 = 0 \Rightarrow 1988 - b + 8891 = 0 \Rightarrow b = 10879$

✅ Final Answer:

$b = \pm 10879$

Q2.

Given that the equation in x has at least a real root, find the range of m.

(m^{2} - 1) x^{2} - 2 (m + 2) x + 1 = 0

Solution:
Once you do discriminant >= 0 you will get m >= -5/4.
Note that this range also includes +-1.
And if you put m = +-1, the given equation is no longer quadratic.
So should we exclude +-1 from this computed range?
No.
Why?
Because when we put m = +-1, the equation becomes a linear equation, with one real root.
And that's what the question asks.
So m >= -5/4 is the answer.

Q3.

If the equation in x has real roots, then find the value of a and b.

x^{2} + 2 (1 + a) x + (3 a^{2} + 4 a b + 4 b^{2} + 2) = 0

Solution:
Once you put discriminant >= 0 and re-arrange, you will get:
(a-1)^2 + (a+2b)^2 = 0
=> a = 1 and b = -0.5

Q4.

Mr. A is going to pick three non-zero real numbers and Mr. B is going to arrange the three numbers as the coefficients of a quadratic equation(ax^2 + bx + c = 0). Mr. A wins the game if and only if the resulting equation has two distinct rational solutions. Who has a winning strategy?

Solution:

Mr. A can force a win.
Here’s a concrete strategy that always succeeds, no matter how Mr. B permutes the three numbers.

1 A's move

Pick any three non‑zero real numbers whose sum is 0 and such that no one of them equals −2 times another (to avoid a double root).
Example: $4,\;1,\;-5$ .

2  Why this choice works for every permutation

Let the three numbers be $x,\;y,\;z$ with

$x+y+z = 0\quad\text{and}\quad x,y,z\neq 0.$

Whatever permutation B chooses becomes the coefficients $(a,b,c)$ of

$a\,t^{2}+b\,t+c=0,\qquad a,b,c\in\{x,y,z\}.$

Because $c = -(a+b)$ , the discriminant is

$\Delta \;=\; b^{2}-4ac \;=\; b^{2}-4a\bigl(-(a+b)\bigr) \;=\; b^{2}+4a^{2}+4ab \;=\; (\,2a+b\,)^{2}.$

Key facts

$(2a+b)^{2}$ is always a perfect square, so the square‑root term in the quadratic formula is rational.
$\Delta > 0$ provided $2a+b\neq 0$ . That’s why A avoids any pair with ratio $-2$ .
(With the example $4,1,-5$ we get squares $9,36,81$ in the six permutations.)

Hence every arrangement gives two distinct roots

$t=\frac{-b\pm\sqrt{\Delta}}{2a} =\frac{-b\pm(2a+b)}{2a},$

a ratio of integers → rational.

3  Conclusion

By picking numbers that sum to zero (and not in the $-2:1$ ratio), Mr. A guarantees that all six possible quadratics have two distinct rational solutions.
Since Mr. B cannot spoil this outcome with any permutation, Mr. A has the winning strategy.

Q5.

a, b, c are three distinct non-zero real numbers. Prove that the following three equations

$ax^2 + 2bx + c = 0,\quad bx^2 + 2cx + a = 0,\quad \text{and} \quad cx^2 + 2ax + b = 0$

cannot all have two equal real roots.

Solution:

Let's prove by contradiction. Let's say each of them has 2 equal real roots. Hence their discriminant is 0.
So
4b^2 = 4ac, 4a^2 = 4bc and 4c^2 = 4ab
=> b^2 = ac and a^2 = bc
=> ac = (a^2/c)^2
=> a^3 = c^3
Since a and c are real,
=> a = c
But that's contradiction since the question says a,b,c are distinct.
Hence proved.

Q6.

If $x^2 + x + 1 = 0$ , find the value of $x^{1999} + x^{2000}$ .
Solution:
Note that this equation gives imaginary cube roots of unity.
x^2 = -1 - x
x^3 = -x - x^2 = 1
Since x^3 is 1
=> x^1999 = x and x^2000 = x^2
So x + x^2 = -1 = Answer.

Q7.

For $x^2 + 2x + 5$ to be a factor of $x^4 + px^2 + q$ , find the values of p and q.

Solution:
Let (x^2 + 2x + 5)(x^2 + bx + c) = x^4 + px^2 + q
Now equate coefficients for each power of x and you will get the answer.
p = 6, q = 25.

Q8.

If $a + b + c = 0$ , find

$\frac{b^{2} + c^{2} + a^{2}}{b^{2} - c a} .$ Solution:
b^2 = a^2 + c^2 + 2ac
Numerator = 2(a^2 + c^2 + ac)
Denominator = a^2 + c^2 + ac
Answer: 2

Q9.

For how many real values of a will

$x^2 + 2ax + 2008 = 0$

have two integer roots?

Solution:

Step 1: General Form

The roots of the quadratic equation $x^2 + 2ax + 2008 = 0$ are:

$x = \frac{-2a \pm \sqrt{(2a)^2 - 4(1)(2008)}}{2} \Rightarrow x = -a \pm \sqrt{a^2 - 2008}$

So for the roots to be real and integers, the quantity $\sqrt{a^2 - 2008}$ must be an integer. Let:

$\sqrt{a^{2} - 2008} = k, for some integer k \geq 0 \Rightarrow a^{2} - 2008 = k^{2} \Rightarrow a^{2} - k^{2} = 2008 \Rightarrow (a - k) (a + k) = 2008$

$(a - k)(a + k) = 2008$

Let’s find all integer factor pairs $(m, n)$ such that $m \cdot n = 2008$ , and then compute $a = \frac{m + n}{2}$ for each.

Step 1: Factorization of 2008

$2008 = 2^3 \times 251$

Number of positive divisors = $(3+1)(1+1) = 8$

So, the positive factor pairs $(a - k, a + k)$ are:

$(1, 2008),\ (2, 1004),\ (4, 502),\ (8, 251)$

We only need pairs $(m, n)$ such that $m < n$ , since the equation is symmetric.

Step 2: Compute $a = \frac{m + n}{2}$

Now for each pair $(m, n)$ , compute:

$a = \frac{m + n}{2}$

$(1, 2008) \Rightarrow a = \frac{1 + 2008}{2} = \frac{2009}{2} = 1004.5$ ❌ Not valid (not real value of a that gives integer roots)
$(2, 1004) \Rightarrow a = \frac{2 + 1004}{2} = \frac{1006}{2} = 503$ ✅
$(4, 502) \Rightarrow a = \frac{4 + 502}{2} = \frac{506}{2} = 253$ ✅
$(8, 251) \Rightarrow a = \frac{8 + 251}{2} = \frac{259}{2} = 129.5$ ❌ Not valid

So only two of the four positive factor pairs give valid real values of $a$ such that the roots are integers.

Step 3: Consider Negative Factor Pairs

Because $a^2 - k^2 = 2008$ , negative factor pairs will also work:

$(-1, -2008),\ (-2, -1004),\ (-4, -502),\ (-8, -251)$

Repeat same for these:

$(-1, -2008) \Rightarrow a = \frac{-1 + (-2008)}{2} = \frac{-2009}{2} = -1004.5$ ❌
$(-2, -1004) \Rightarrow a = \frac{-1006}{2} = -503$ ✅
$(-4, -502) \Rightarrow a = \frac{-506}{2} = -253$ ✅
$(-8, -251) \Rightarrow a = \frac{-259}{2} = -129.5$ ❌

✅ Final Set of Valid $a$ Values:

$a = \pm 503,\ \pm 253$

So the 4 distinct real values of $a$ are:

$- 503, - 253, 253, 503$

Q10.

If $x, y$ are positive real numbers satisfying the system of equations

$x^2 + y \sqrt{xy} = 336, \quad y^2 + x \sqrt{xy} = 112,$

then $x + y$ equals

Solution:

We are given:

x^2 + y\sqrt{xy} = 336 \tag{1}

Dividing both sides of (1) by $\sqrt{x}$ :

x\sqrt{x} + y\sqrt{y} = \frac{336}{\sqrt{x}} \tag{2}

We are also given:

y^2 + x\sqrt{xy} = 112 \tag{3}

Dividing both sides of (3) by $\sqrt{y}$ :

y\sqrt{y} + x\sqrt{x} = \frac{112}{\sqrt{y}} \tag{4}

From equations (2) and (4), we observe that both left-hand sides are the same (just terms rearranged), so:

\frac{336}{\sqrt{x}} = \frac{112}{\sqrt{y}} \tag{5}

Cross-multiplying:

336\sqrt{y} = 112\sqrt{x}

Divide both sides by 112:

3\sqrt{y} = \sqrt{x} \tag{6}

Squaring both sides:

9y = x \tag{7}

Now substitute equations (6) and (7) into equation (2):

Recall:

x\sqrt{x} + y\sqrt{y} = \frac{336}{\sqrt{x}}

From (7), $x = 9y$ , and from (6), $\sqrt{x} = 3\sqrt{y}$ , so:

Left-hand side:

x\sqrt{x} = 9y \cdot 3\sqrt{y} = 27y\sqrt{y}

y\sqrt{y} = y\sqrt{y}

\text{Total LHS} = 27y\sqrt{y} + y\sqrt{y} = 28y\sqrt{y}

Right-hand side:

\frac{336}{\sqrt{x}} = \frac{336}{3\sqrt{y}} = \frac{112}{\sqrt{y}}

So:

28y\sqrt{y} = \frac{112}{\sqrt{y}}

Multiply both sides by $\sqrt{y}$ :

28y y = 112 \Rightarrow 28y^2 = 112 \Rightarrow y^2 = 4 \Rightarrow y = 2 \quad \text{(since } y > 0\text{)}

Using equation (7):

x = 9y = 9 \cdot 2 = 18

Thus,

x + y = 18 + 2 = 20

Q11.

a, b, c are positive integers such that

$a^2 + 2b^2 - 2bc = 100$ and $2ab - c^2 = 100$ . Then $\frac{a + b}{c}$ is"
Solution:
Equating both you will get:
(a-b)^2 + (b-c)^2 = 0
=> a = b = c
Answer = 2

Q12.

"When $x$ is real, the greatest possible value of $10^x - 100^x$ is"
Solution:
Let y = 10^x then:
y - y^2 is a quadratic.
Its graph will be downward parabola.
Highest point will be at the vertex.
Vertex for a quadratic is -b/2a
Here b = 1, a = -1
Vertex = 1/2
Put y = 1/2 to get max value of 1/2 - 1/4 = 1/4 = answer.

Q13. Find integers 'a' and 'b' such that

$(x^2 - x - 1)$ divides $ax^{17} + bx^{16} + 1$ .

Solution:
For a polynomial to divide another, all its roots must be roots of the other one.
Let's say one of those roots is p.
=> p^2 - p - 1 = 0
=> p^2 = p + 1________[1]
Also
a.p^17 + b.p^16 + 1 = 0______[2]

Using [1], we can reduce higher powers of p to linear.
While doing that you will realize that coefficients turn out to follow Fibonacci.
i.e.
p^n = F_n.x + F_(n-1)
Using that compute p^16 and p^17 and put in 2 to get:

$P(x) = a(F_{17}x + F_{16}) + b(F_{16}x + F_{15}) + 1$ $= (aF_{17} + bF_{16})x + (aF_{16} + bF_{15} + 1).$

For this to vanish identically we need

$\begin{cases} aF_{17} + bF_{16} = 0, \\ aF_{16} + bF_{15} + 1 = 0. \end{cases}$

Plugging the numerical values gives the linear system

$\begin{cases} 1597a + 987b = 0, \\ 987a + 610b = -1. \end{cases}$

Solving this will give, a = 987, b = -1597

Q14.

Find the real points (x, y) satisfying

$3x^2 + 3y^2 - 4xy + 10x - 10y + 10 = 0$ .

Solution:

Solve it as a plain quadratic

Start from

$3x^{2}+3y^{2}-4xy+10x-10y+10=0. \tag{1}$

1. View the equation as a quadratic in  $x$

Rewrite (1) collecting the $x$ -terms:

$3x^{2}\;+\;(-4y+10)\,x\;+\;\bigl(3y^{2}-10y+10\bigr)=0.$

For a fixed real $y$ , this is a quadratic in $x$ .
Real solutions require the discriminant to be non‑negative:

$\Delta_x \;=\;(-4y+10)^{2}-4\cdot3\bigl(3y^{2}-10y+10\bigr)\;\ge 0.$

Compute the discriminant:

$\begin{aligned} (-4y+10)^{2}&=16y^{2}-80y+100,\\[2pt] 4\cdot3\bigl(3y^{2}-10y+10\bigr)&=12\bigl(3y^{2}-10y+10\bigr)=36y^{2}-120y+120. \end{aligned}$

Hence

$\Delta_x=(16y^{2}-80y+100)-(36y^{2}-120y+120) =-20y^{2}+40y-20 =-20\bigl(y^{2}-2y+1\bigr) =-20\,(y-1)^{2}.$

Because $-20<0$ , the only way for $\Delta_x\ge0$ is

$(y-1)^{2}=0\quad\Longrightarrow\quad y=1.$

2. Substitute $y=1$ back into the equation

Put $y=1$ in (1):

$3x^{2}+3(1)^{2}-4x(1)+10x-10(1)+10=0 \;\;\Longrightarrow\;\; 3x^{2}+6x+3=0.$

Divide by $3$ :

$x^{2}+2x+1=0\quad\Longrightarrow\quad(x+1)^{2}=0,$

so $x=-1$ .

3. Conclusion

The only real point that satisfies (1) is

$\boxed{(-1,\,1)}.$

(Had we treated the equation instead as a quadratic in $y$ , the same reasoning would force $x=-1$ and then give $y=1$ ; either way the result is identical.)

Q15.

Solve for x, y and z; if xy + x + y = 23, yz + y + z = 31, zx + z + x = 47.

Solution:
Add 1 to each equation:
(x+1)(y+1) = 24
(y+1)(z+1) = 32
(x+1)(z+1) = 48
Let x+1 = a, y+1 = b, z+1 = c
ab = 24
bc = 32
ac = 48
ac.bc = 32.48 = ab.c^2 => c^2 = 64 => c = +-8
c = 8 gives a = 6, b = 4 => x = 5, y = 3, z = 7
c = -8 gives a = -6, b = -4 => x = -7, y = -5, z = -9

Q16

a, b are the real roots of the equation

$x^2 - px + q = 0$ . Find the number of pairs $(p, q)$ such that the quadratic equation with roots $a^2, b^2$ is still $x^2 - px + q = 0$ .

Solution

1. Roots and Symmetric Sums

Given: $a, b$ are roots of $x^2 - px + q = 0$ .
- So $a + b = p$
- $ab = q$

We require that the quadratic with roots $a^2, b^2$ is also $x^2 - px + q = 0$ .

For the new quadratic with roots $a^2, b^2$ :

Sum of roots: $a^2 + b^2$
Product of roots: $a^2 b^2$

So, we want:

$a^2 + b^2 = p$
$a^2 b^2 = q$

But $a^2 + b^2 = (a + b)^2 - 2ab = p^2 - 2q$ .

$a^2 b^2 = (ab)^2 = q^2$

Thus,

$a^2 + b^2 = p^2 - 2q = p$
$a^2 b^2 = q^2 = q$

So,

From (2): $q^2 - q = 0 \implies q(q - 1) = 0$ , so $q = 0$ or $q = 1$ .

We will examine both cases.

2. Case Analysis

Case 1: $q = 0$

From (1): $p^2 - 2q = p \implies p^2 - p = 0 \Rightarrow p(p - 1) = 0$

So $p = 0$ or $p = 1$ .

Now, check if the roots are real.

Case 2: $q = 1$

From (1): $p^2 - 2q = p \implies p^2 - p - 2 = 0 \implies (p - 2)(p + 1) = 0$

So $p = 2$ or $p = -1$ .

Again, check if the roots are real.

3. Checking Nature of Roots

The original equation $x^2 - px + q = 0$ will have real roots if its discriminant $D \geq 0$ :

D = p^2 - 4q \geq 0

For each (p, q):

(0, 0): $D = 0^2 - 0 = 0$ (real roots)
(1, 0): $D = 1^2 - 0 = 1$ (real roots)
(2, 1): $D = 4 - 4 = 0$ (real roots)
(-1, 1): $D = 1 - 4 = -3$ (NOT real roots)

So only 3 pairs yield real roots:

(0, 0)
(1, 0)
(2, 1)

Final Answer

There are $\boxed{3}$ pairs $(p, q)$ such that the quadratic equation with roots $a^2, b^2$ is also $x^2 - px + q = 0$ , where $a, b$ are real roots of the original equation.

These pairs are:

(0, 0)
(1, 0)
(2, 1)

Q17.
Done earlier in examples.

Q18.

Given that the real numbers $s$ , $t$ satisfy

$19s^2 + 99s + 1 = 0,\quad t^2 + 99t + 19 = 0,\quad \text{and } st \ne 1,$

find the value of

$\frac{s t + 4 s + 1}{t} .$ Solution:

Changing the second equality in the form of

$19\left(\frac{1}{t}\right)^2 + 99\left(\frac{1}{t}\right) + 1 = 0,$

it follows that $s$ and $\frac{1}{t}$ both are roots of the equation

$19x^2 + 99x + 1 = 0.$

Therefore, by Viete's Theorem,

$s + \frac{1}{t} = -\frac{99}{19}, \quad \text{and} \quad \frac{s}{t} = \frac{1}{19}.$ $\therefore \frac{st + 4s + 1}{t} = s + \frac{1}{t} + 4\cdot\frac{s}{t} = -\frac{99}{19} + \frac{4}{19} = -5$

Q19.

Given that a = 8 − b and c² = ab − 16, prove that a = b.
Solution:
Substitute a in second equation to get:
c^2 + (b-4)^2 = 0
=> c = 0, b = 4
=> a = 8 - 4 = 4
H.P.
Note: This solution assumes that a,b,c are real. Else there are imaginary solutions where a != b.
For e.g. a = 4 + i and b = 4 - i.
And generally, c = +-i(b - 4). We can pick any b and use that to compute c and a.

Q20. Given that a, b are roots of the equation x² – 7x + 8 = 0, where a > b. Find the value of

$\frac{2}{a} + 3b^2$ without solving the equation.:

Solution:

Let α and β be the two roots of

$x^{2}-7x+8=0,\qquad\text{with } \alpha>\beta .$

1. Use Vieta’s relations

$\alpha+\beta = 7 ,\qquad \alpha\beta = 8 .$

2. Rewrite the required expression

$S \;=\; \frac{2}{\alpha}+3\beta^{2}.$

Because $\alpha\beta = 8$ ,

$\frac{1}{\alpha}=\frac{\beta}{\alpha\beta}= \frac{\beta}{8}\quad\Longrightarrow\quad \frac{2}{\alpha}= \frac{\beta}{4}.$

Hence

$S = \frac{\beta}{4}+3\beta^{2}. \tag{1}$

3. Eliminate the square of β

Since β is a root, it satisfies its own equation:

$\beta^{2}-7\beta+8 = 0\;\;\Longrightarrow\;\; \beta^{2}=7\beta-8 .$

Substitute this into (1):

$\begin{aligned} S &=\frac{\beta}{4}+3(7\beta-8)\\ &=\frac{\beta}{4}+21\beta-24\\ &=\frac{85\beta}{4}-24. \tag{2} \end{aligned}$

4. Express β in terms of S and eliminate β

From (2): $\beta = \dfrac{4S+96}{85}$ .

Insert this form of β back into the quadratic it satisfies:

$\left(\frac{4S+96}{85}\right)^{2}-7\left(\frac{4S+96}{85}\right)+8=0.$

Multiply by $85^{2}$ and simplify:

$16S^{2}-1612S+9896=0 \;\;\Longrightarrow\;\; 4S^{2}-403S+2474=0.$

5. Solve for S

$S=\frac{403\pm\sqrt{403^{2}-4\cdot4\cdot2474}}{8} =\frac{403\pm\sqrt{85^{2}\cdot17}}{8} =\frac{403\pm85\sqrt{17}}{8}.$

Because $\alpha>\beta$ , β is the smaller root, giving the smaller of the two S‑values (the “−” sign):

$S = \frac{403 - 85 \sqrt{17}}{8} (\approx 6.57) .$

Q21. Both roots of the quadratic equation x² – 63x + k = 0 are prime numbers. The number of possible values of k is:
Solution:
Let the roots be a,b.
a + b = 63
a,b are prime => a,b are positive integers.
2 + 61 = 63 gives one pair(2,61) and k = 122.
Since a,b are prime and all primes except 2 are odd.
So if we add 2 primes and both are odd we will get an even number.
But here the sum is odd. So one of them as to be even.
So only answer is the one we found.
Number of possible values of k = 1.

Q22.

Find the sum of all possible values of a such that the following equation has real root in x:

$(x - a)^{2} + (x^{2} - 3 x + 2)^{2} = 0.$
Solution:
For the LHS to be zero, both squares have to be 0.
The second term will be zero only for x = 1 and 2.
So for the first term to be zero a will have to be 1 or 2.
Answer: 1 + 2 = 3

Q23:

Let p be a real number such that the equation $2y^2 - 8y = p$ has only one solution.
Then find the value of $p + 64$ .

Solution:
Discriminant has to be 0.
64 + 8p = 0 => p = -8 => p + 64 = 56.

Q24:

Let

$a, b$ and $c$ be the lengths of the three sides of a triangle. Suppose $a$ and $b$ are the roots of the equation

$x^{2} + 4 (c + 2) = (c + 4) x,$ Solution:

Since $a, b$ are the roots of the equation $x^2 - (c + 4)x + 4(c + 2) = 0$ , it follows that

$a + b = c + 4 \quad \text{and} \quad ab = 4(c + 2)$

Then

$a^2 + b^2 = (a + b)^2 - 2ab = (c + 4)^2 - 8(c + 2)$ $\Rightarrow a^2 + b^2 = c^2$

Hence the triangle is right-angled, and $x = 90$ .

Q25:

How many ordered pairs of integers (x, y) satisfy the equation

x² + y² = 2(x + y) + xy?

Solution:

x² + y² = 2(x + y) + xy

Key steps in the solution:

Rewriting the equation as a quadratic in $x$ :
$x^2 - x(2 + y) + y^2 - 2y = 0$
Discriminant of the quadratic:
$\text{Discriminant} = (2 + y)^2 - 4(y^2 - 2y) = 16 - 3(y - 2)^2$
To ensure integer solutions, this must be a perfect square.
Solving the inequality:
$16 - 3(y - 2)^2 \geq 0 \Rightarrow |y - 2| < \frac{4}{\sqrt{3}} < 3$
So possible integer values for $y$ : 0, 1, 2, 3, 4.
Filtering values where discriminant is a perfect square: Only $y = 0, 2, 4$ qualify.
Solving for each valid $y$ :
- $y = 0$ : $x = 0, 2$
- $y = 2$ : $x = 0, 4$
- $y = 4$ : $x = 2, 4$
Final integer solutions:
$(0,0), (2,0), (0,2), (4,2), (2,4), (4,4)$

Total number of solutions: 6 ordered pairs.

Q26.

Let n be a positive integer such that one of the roots of the quadratic equation

4x² – (4√3 + 4)x + √3n – 24 = 0
is an integer. Find the value of n.

Solution:

$4x^2 - (4\sqrt{3} + 4)x + \sqrt{3}n - 24 = 0$

Summary of the key steps:

Assume an integer root $\alpha$ , substitute into the equation:
$4\alpha^2 - (4\sqrt{3} + 4)\alpha + \sqrt{3}n - 24 = 0 \Rightarrow 4\alpha^2 - 4\alpha - 24 = \sqrt{3}(4\alpha - n)$
Since $\sqrt{3}$ is irrational, equate both rational and irrational parts:
- $4\alpha^2 - 4\alpha - 24 = 0$
- $4\alpha - n = 0 \Rightarrow \alpha = \frac{n}{4}$
Substituting $\alpha = \frac{n}{4}$ into the rational equation:
$4\left(\frac{n}{4}\right)^2 - 4\left(\frac{n}{4}\right) - 24 = 0 \Rightarrow n^2 - 4n - 96 = 0 \Rightarrow (n - 12)(n + 8) = 0$
Since $n$ is a positive integer, the solution is:
$\boxed{n = 12}$

Q27.Suppose that the two roots of the equation

$\frac{1}{x^2 - 10x - 29} + \frac{1}{x^2 - 10x - 45} - \frac{2}{x^2 - 10x - 69} = 0$

are α and β. Find the value of (−αβ).

Solution:

Let

$y = x^2 - 10x - 29$ . Then $x^2 - 10x - 45 = y - 16$ , $x^2 - 10x - 69 = y - 40$ . Thus

\frac{1}{y} + \frac{1}{y - 16} - \frac{2}{y - 40} = 0.

Therefore $y = 10$ . So $y = x^2 - 10x - 29 = 10$ . Hence $x^2 - 10x - 39 = 0$ . Therefore, $-\alpha\beta = 39$ .

Q28:

Find total number of pairs of (x, y) where x, y belongs to integer that satisfy the equation
x + y = x² - xy + y².
Solution:
Rearrange to make quadratic in x:
x^2 - x(y+1) + y^2 - y = 0
Discriminant >= 0
=> (y+1)^2 - 4(y^2 - y) >= 0
=> -3y^2 + 6y + 1 >= 0
=> 3y^2 - 6y - 1 <= 0
=> 3[y^2 - 2y] <= 1
=> 3[(y-1)^2 - 1] <= 1
=> (y-1)^2 <= 1 + 1/3 = 4/3
=> |y-1| <= 2/sqrt(3)
=> -2/sqrt(3)<= y-1 <= 2/sqrt(3)
=> -1... <= y-1 <= 1...
=> -0... <= y <= 2...
Since y is integer
y = 0,1,2

Since x is integer, Discriminant should be perfect square.
Discriminant = (y+1)^2 - 4(y^2 - y) = D
y = 0 => D = 1 Ok
y = 1 => D = 4 Ok
y = 2 => D = 9 - 4(2) = 1 Ok

Now, let's find values of x using
x^2 - x(y+1) + y^2 - y = 0
y = 0
=> x^2 - x = 0 => x = 0,1
y = 1
=> x^2 - 2x = 0 => x = 0,2
y = 2
=> x^2 - 3x + 2 = 0 => x = 1,2
So total 6 pairs for (x,y):
(0,0),(1,0),(0,1),(2,1),(1,2),(2,2)

Q29.

Let a and b be two integers. Suppose that √(7 - 4√3) is a root of the equation x² + ax + b = 0. Find the value of b - a.

Solution:
First note that sqrt(7 - 4.sqrt(3)) = 2 - sqrt(3).
If a quadratic has an irrational root but integer coefficients, then the other root will be conjugate of the first one.
Why?
Let's say p + k.sqrt(q) is one root.
Since sum of roots is integer, so the other root will be like t - k.sqrt(q) so that the irrational part vanishes.
Now the product of roots is also integer.
Product of roots = (p + k.sqrt(q))(t - k.sqrt(q)) = pt - k^2.q + k.sqrt(q)(t - p)
For it to be integer, the irrational part should vanish => t = p.
So the other root is p - k.sqrt(q).

Now, coming back to the question at hand.
Since the coefficients are integer and one root is 2 - sqrt(3) the other one will be 2 + sqrt(3).
-a = sum of roots = 4 => a = -4
b = product of roots = 4 - 3 = 1
b-a = 5 = Answer.

Q30.

Let $p$ be an integer such that both roots of the equation

$5x^2 - 5px + (66p - 1) = 0$

are positive integers. Find the value of $p$ .

Solution:
Let the roots be r,s.
r + s = p
rs = (66p - 1)/5 = [66(r+s)- 1 ]/5
5rs = 66r + 66s - 1
5rs - 66r - 66s = - 1
Using rectangle completion trick:
5*5rs - 5*66r - 5*66s = -5
5r(5s - 66) - 66(5s) = -5
5r(5s - 66) - 66(5s - 66) - 66^2 = -5
(5r - 66)(5s - 66) = 66^2 - 5 = 4351

Let's factor 4351.
Last digit is 1.
So last digits of factors would be:
1,1 or 3,7 or 9,9
Check divisibility by 3,7,11,13,17 - all fail.
19 succeeds and the other factor is 229.
(5r - 66)(5s - 66) = 19*229
Solving this we get the roots as :
17,59
p = 17 + 59 = 76

Q31.

Suppose $a$ and $b$ are the roots of $x^2 + x \sin \alpha + 1 = 0$ while $c$ and $d$ are the roots of the equation
$x^2 + x \cos \alpha - 1 = 0$ . Find the value of

$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2}.$

Solution:
p = sin(alpha), q = cos(alpha)
a + b = -p ab = 1
c + d = -q cd = -1

1/a^2 + 1/b^2 = (a^2 + b^2)/(a^2.b^2) = p^2 - 2/(ab)^2 = p^2 - 2
1/c^2 + 1/d^2 = (c^2 + d^2)/(c^2.d^2) = q^2 + 2/(cd)^2 = q^2 + 2
Add to get p^2 + q^2 = 1 = answer.

Q32.

The zeroes of the function $f(x) = x^2 - ax + 2a$ are integers. What is the sum of all possible values of $a$ ?

Solution:
Let the roots be p,q.
p + q = a
pq = 2a = 2(p+q)
pq -2p - 2q = 0
Complete the rectangle
p(q-2) -2(q-2) = 4
(p-2)(q-2) = 4
Possible factors are 1,4 2,2 4,1 -1,-4 -2,-2 -4,-1
Solving these and doing a = p+q gives 4 unique values of a:
9,8,0,-1
For each of these the discriminant is also perfect square.
Add all to get 16 = answer.

Q33.

Let $a, b,$ and $c$ be three distinct one-digit numbers. The maximum value of the sum of the roots of the equation

$(x - a)(x - b) + (x - b)(x - c) = 0$

is $k$ . What is the value of $2k$ ?

Solution:
Factorize further to get:
(x-b)(2x - a - c) = 0
x = b, (a+c)/2
k = b + (a+c)/2
2k = 2b + a + c
To maximize this, put
b = 9 and either of a and c to 8 and other one 7.
18 + 8 + 7 = 33 = answer.

Q34.

There are exactly N distinct rational numbers k such that |k| < 200 and 5x² + kx + 12 = 0 has at least one integer solution for x. What is N?

Solution:
Let n be the integer root.
5n^2 + nk + 12 = 0
=> k = -(12 + 5n^2)/n = -(12/n + 5n)
|k| < 200 => |5n + 12/n| < 200
=> -200 < (5n + 12/n) < 200

Claim: Each distinct value of n gives a distinct value of k. Hence N = number of distinct 'n'.
Proof by contradiction:
Let's say 2 distinct value of n(n1,n2) give same k.
=> 5n1 + 12/n1 = 5n2 + 12/n2
=> 12(1/n1 - 1/n2) - 5(n2 - n1) = 0
=> (n2 - n1)[12/n1n2 - 5] = 0
=> (n2 - n1)[12 - 5n1n2] = 0
Either n2 = n1(not true since we chose them distinct)
Or 12 = 5n1n2 (not true since 5 and 12 are co-prime).
H.P.

Case 1:
5n^2 - 200n + 12 < 0
Quadratic in 'n', graph opens upwards, so will be negative between the roots.
n = 200 +- sqrt(40000 - 240)/10
= 20 +- sqrt(400 - 2.4) = 20 +- 19.sth = 39.sth, 0.sth
Since n is integer the values will be 1 to 39.

Case 2:
5n^2 + 200n + 12 > 0
This above inequality is wrong. Why?
Because I have multiplied with n but n is negative here. So the sign must flip.
So it should become:
5n^2 + 200n + 12 < 0
Quadratic in 'n', graph opens upwards, so will be positive when 'n' is between the roots.
n = -200 +- sqrt(40000 - 240)/10
= -20 +- sqrt(400 - 2.4) = -20 +- 19.sth = -39.sth, -0.sth
=> -39 <= n <= -1

So total value of n = 78 = total values of k.
Q35.

Let $a$ and $b$ be the roots of $x^2 + 2000x + 1 = 0$ and let $c$ and $d$ be the roots of $x^2 - 2008x + 1 = 0$ . Now $P$ is defined as

$P = (a + c)(b + c)(a - d)(b - d)$

Then find the sum of all the digits of $P$ .

Solution:
(a + c)(b + c) = ab + c^2 + c(a+b) = 1 - 2000c + c^2 = 8c
(a - d)(b - d) = ab + d^2 - d(a+b) = 1 - 2000d - d^2 = 4008d
Multiply to get 32064cd = 32064 = P
Since c,d are roots of the second equation:
c^2 - 2008c + 1 = 0
d^2 - 2008d + 1 = 0
Sum of digits of P = 15 = Answer.

Q36.

Find the value of the smallest positive integer $m$ such that the equation

$x^2 + 2(m + 5)x + (100m + 9) = 0$

has only integer solutions.

Solution:
D = perfect square => 4(m+5)^2 - 4(100m + 9) = p.s.
Factor out 4 since that's already p.s.
(m+5)^2 - (100m + 9) >= 0 and p.s.
m^2 - 90m + 16 >= 0 and p.s.
It's given that m is positive integer.
As long as m <= 89 this LHS expression is negative.
For e.g. at m = 89
89(89 - 90+ 16 < 0
So the smallest m at which LHS > 0 is m = 90.
And also at m = 90 it's 16 which is p.s.
Even though we have found the answer here is a more detailed solution.

m^2 + 10m - 100m + 16 = k^2
(m - 45)^2 - 2025 + 16 = k^2
(m - 45)^2 - k^2 = 2009
Using integer square solution method:
(| m - 45| + |k|) (| m - 45| - |k|) = 2009
Factors of 2009 = 7*7*41
=> |m - 45| = 1005,147,45
Smallest positive m will be obtained by:
m - 45 = +-45
=> m = 90 = answer

Q37.

Given that (m − 2) is a positive integer and it is also a factor of 3m² − 2m + 10. Find the sum of all such values of m.

Solution:

$\frac{3m^2 - 2m + 10}{m - 2} = 3m + 4 + \frac{18}{m - 2}$

is an integer. Thus, $m - 2$ is a factor of 18. Hence, value of $m - 2 = 1, 2, 3, 6, 9, 18$ . Thus,

$m = 3, 4, 5, 8, 11, 20.$

The required sum is 51.

Q38.

Let

$a$ and $b$ be the roots of the equation $x^2 - mx + 2 = 0$ . Suppose that $a + \frac{1}{b}$ and $b + \frac{1}{a}$ are the roots of the equation $x^2 - px + q = 0$ . What is $2q$ ?

Solution:

In a quadratic equation of the form

$x^2 + bx + c = 0$ , the product of the roots is $c$ (Vieta’s formulas).
Using this property, we have that $ab = 2$ and

$q = \left(a + \frac{1}{b}\right)\left(b + \frac{1}{a}\right) = \frac{ab + 1}{b} \cdot \frac{ab + 1}{a} = \frac{(ab + 1)^2}{ab} = \frac{(2 + 1)^2}{2} = \frac{9}{2}$

Hence,

$2 q = 9$
Q39.

The quadratic equation

$x^2 + mx + n = 0$ has roots that are twice those of $x^2 + px + m = 0$ , and none of $m, n, p$ is zero. What is the value of $\frac{n}{p}$ ?

Solution:

Let

$x^2 + px + m = 0$ have roots $a$ and $b$ . Then

$x^2 + px + m = (x - a)(x - b) = x^2 - (a + b)x + ab,$

so $p = -(a + b)$ and $m = ab$ .
Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so

$x^2 + mx + n = (x - 2a)(x - 2b) = x^2 - 2(a + b)x + 4ab,$

and $m = -2(a + b)$ , $n = 4ab$ .
Thus,

$\frac{n}{p} = \frac{4 a b}{- (a + b)} = \frac{4 m}{\frac{m}{2}} = 8$
Q40.

What is the negative of the sum of the reciprocals of the roots of the equation

$\frac{2003}{2004} x + 1 + \frac{1}{x} = 0 ?$ Solution:

Multiplying both sides of the equation

$\frac{2003}{2004}x + 1 + \frac{1}{x} = 0$

by $x$ , we get:

$\frac{2003}{2004}x^2 + x + 1 = 0$

Let the roots be $a$ and $b$ .

The question asks for the negative of the sum of the reciprocals of the roots:

$-\left(\frac{1}{a} + \frac{1}{b}\right) = -\left(\frac{a + b}{ab}\right)$

Using Vieta’s formulas:

$a + b = -\frac{1}{\frac{2003}{2004}} = -\frac{2004}{2003}$
$ab = \frac{1}{\frac{2003}{2004}} = \frac{2004}{2003}$

So:

$\frac{a + b}{ab} = \frac{-\frac{2004}{2003}}{\frac{2004}{2003}} = -1$

Therefore, the negative of this is:

$- (- 1) = 1$
Q42. Suppose that

$a$ and $b$ are nonzero real numbers, and that the equation $x^2 + ax + b = 0$ has solutions $a$ and $b$ . Find the value of $(a - b)$ .

Solution:

Since $(x - a)(x - b) = x^2 - (a + b)x + ab = x^2 + ax + b = 0$ ,
it follows by comparing coefficients that $-a - b = a$ and that $ab = b$ .
Since $b$ is non-zero, $a = 1$ , and

$-1 - b = 1 \Rightarrow b = -2$

Thus,

$a - b = 3$ Q43.

Let

$a, b,$ and $c$ be real numbers such that
$a - 7b + 8c = 4$ and $8a + 4b - c = 7$ .
Then $a^2 - b^2 + c^2$ is ___?

Solution:

Rearranging the given equations:

$a + 8c = 7b + 4$
$8a - c = 7 - 4b$

Squaring both:

$a^2 + 16ac + 64c^2 = 49b^2 + 56b + 16$
$64a^2 - 16ac + c^2 = 16b^2 - 56b + 49$

Adding these two equations:

$a^2 + 64a^2 + 16ac - 16ac + 64c^2 + c^2 = 49b^2 + 16b^2 + 56b - 56b + 16 + 49$

Simplifies to:

$65a^2 + 65c^2 = 65b^2 + 65$

Dividing both sides by 65:

$a^2 + c^2 = b^2 + 1$

Therefore:

$a^{2} - b^{2} + c^{2} = 1$

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